Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation 5x-9=3x+1
-
Equation -9*(8-9x)=4x+5
-
Equation (x^2+4)/x=0
-
Equation x^2+4x+2=0
- Express {x} in terms of y where:
- sqrt(x)+sqrt(y)=69,49
- (x^2+y^2)×(x+y-3)=2xy
- sqrt(1-x)^2+(4-y)^2=0
- lnx+sin(y/x+2x)=4
- Graphing y =:
- 2x^3+x^2
-
Identical expressions
- two x^ three +x^2
- 2x cubed plus x squared
- two x to the power of three plus x squared
- 2x3+x2
- 2x³+x²
- 2x to the power of 3+x to the power of 2
-
Similar expressions
- 2x^3-x^2
2x^3+x^2 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$2 x^{3} + x^{2} = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(2 x^{2} + x\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$2 x^{2} + x = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 2$$
$$b = 1$$
$$c = 0$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = 0$$
$$x_{3} = - \frac{1}{2}$$
The final answer for 2*x^3 + x^2 = 0:
$$x_{1} = 0$$
$$x_{2} = 0$$
$$x_{3} = - \frac{1}{2}$$
$$2 x^{3} + x^{2} = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(2 x^{2} + x\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$2 x^{2} + x = 0$$
This equation is of the form
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 2$$
$$b = 1$$
$$c = 0$$
, then
D = b^2 - 4 * a * c =
(1)^2 - 4 * (2) * (0) = 1
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = 0$$
$$x_{3} = - \frac{1}{2}$$
The final answer for 2*x^3 + x^2 = 0:
$$x_{1} = 0$$
$$x_{2} = 0$$
$$x_{3} = - \frac{1}{2}$$
Vieta's Theorem
rewrite the equation
$$2 x^{3} + x^{2} = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} + \frac{x^{2}}{2} = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{1}{2}$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = - \frac{1}{2}$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = 0$$
$$2 x^{3} + x^{2} = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} + \frac{x^{2}}{2} = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{1}{2}$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = - \frac{1}{2}$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = 0$$
Sum and product of roots
[src]
sum
-1/2
$$- \frac{1}{2}$$
=
-1/2
$$- \frac{1}{2}$$
product
0*(-1) ------ 2
$$\frac{\left(-1\right) 0}{2}$$
=
0
$$0$$
0